Question: Ify is floating freely in outer space. The gravitational forces pulling on her body are negligible except for the forces from a nearby asteroid ( ${\vec a}$ ), the nearest planet ( ${\vec p}$ ), and the nearest star ( ${\vec s}$ ). The asteroid, planet, star, and Ify all lie in the same plane. ${\vec{a}} = (2,4)$ ${\vec{p}} = (-1,-3)$ ${\vec{s}} = (8,-2)$ (Forces are given in newtons, $\text{N}$.) What is the magnitude of the net gravitational force acting on Ify?
Explanation: Whenever forces are pulling in different directions, they tend to partially cancel each other out. This canceling effect is exactly what happens when we add vectors. Consider what happens when we add the three vectors presented in the problem. $\vec g = {\vec{a}} + {\vec{p}} + {\vec{s}} = {(2,4)} + {(-1,-3)} + {(8,-2)} = (9, -1)$ The vector $\vec g = (9,-1)$ describes the net gravitational force acting on Ify. We can find the magnitude of $\vec g$ using the Pythagorean theorem. $\begin{aligned} \| \vec g \|^2 &= 9^2 + (-1)^2\\\\ \| \vec g \| &= \sqrt{81 + 1}\\\\ \| \vec g \| &= \sqrt{82}\\\\ \| \vec g \| &\approx 9.1 \text{ N} \end{aligned}$ Finding the direction of $\vec g$ will tell us in what direction Ify is getting pulled. $\vec g$ is pointing in the fourth quadrant with an $x$ -component of $9$ and a $y$ -component of $-1$. We can find the direction of any vector $\vec v$ in the fourth quadrant using the arctangent function and adding $2\pi$. $\begin{aligned} \tan \theta &= \dfrac{ y}{ x}\\\\ \tan \theta &= \dfrac{-1}{9}\\\\ \theta &= \arctan{\left ( -\dfrac{1}{9} \right ) } \\\\ \theta&\approx -0.11\,\text{radians} \end{aligned}$ Adding $2\pi$ to this result gives us $6.2$ radians. The magnitude of the net gravitational force acting on Ify is $9.1$ newtons. Ify is getting pulled in a direction of $6.2$ radians.